Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $y = \dfrac{-10q^2 + 80q}{q^2 - 10q + 16} \div \dfrac{q + 7}{2q - 4} $
Dividing by an expression is the same as multiplying by its inverse. $y = \dfrac{-10q^2 + 80q}{q^2 - 10q + 16} \times \dfrac{2q - 4}{q + 7} $ First factor the quadratic. $y = \dfrac{-10q^2 + 80q}{(q - 2)(q - 8)} \times \dfrac{2q - 4}{q + 7} $ Then factor out any other terms. $y = \dfrac{-10q(q - 8)}{(q - 2)(q - 8)} \times \dfrac{2(q - 2)}{q + 7} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ -10q(q - 8) \times 2(q - 2) } { (q - 2)(q - 8) \times (q + 7) } $ $y = \dfrac{ -20q(q - 8)(q - 2)}{ (q - 2)(q - 8)(q + 7)} $ Notice that $(q - 8)$ and $(q - 2)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -20q(q - 8)\cancel{(q - 2)}}{ \cancel{(q - 2)}(q - 8)(q + 7)} $ We are dividing by $q - 2$ , so $q - 2 \neq 0$ Therefore, $q \neq 2$ $y = \dfrac{ -20q\cancel{(q - 8)}\cancel{(q - 2)}}{ \cancel{(q - 2)}\cancel{(q - 8)}(q + 7)} $ We are dividing by $q - 8$ , so $q - 8 \neq 0$ Therefore, $q \neq 8$ $y = \dfrac{-20q}{q + 7} ; \space q \neq 2 ; \space q \neq 8 $